Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& os, vector<T> a) {
  os << "( ";
  for (T& x : a) os << x << " ";
  os << ")\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T> a) {
  os << "( ";
  for (T x : a) os << x << " ";
  os << ")\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, pair<T, T> a) {
  os << "(" << a.first << " " << a.second << ") ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, complex<T> a) {
  os << "(r=" << a.real() << " i=" << a.imag() << ") ";
  return os;
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long H, L, R;
  cin >> H >> L >> R;
  int n;
  cin >> n;
  vector<pair<long long, long long> > a(n);
  for (auto& A : a) cin >> A.first >> A.second;
  vector<int> s(n + 3);
  for (int i = 1; i <= n; ++i) {
    s[i] = s[i - 1] + a[i - 1].second - a[i - 1].first;
  }
  int q;
  cin >> q;
  while (q--) {
    pair<long long, long long> C;
    cin >> C.first >> C.second;
    int d = 0, c = n - 1, l = n, r = -1;
    while (d <= c) {
      int g = (d + c) / 2;
      long long first = ((C.first - a[g].second) * H + a[g].second * C.second);
      if (first >= L * C.second) {
        l = g;
        c = g - 1;
      } else
        d = g + 1;
    }
    d = 0, c = n - 1;
    while (d <= c) {
      int g = (d + c) / 2;
      long long first = ((C.first - a[g].first) * H + a[g].first * C.second);
      if (first <= R * C.second) {
        r = g;
        d = g + 1;
      } else
        c = g - 1;
    }
    if (l > r) {
      printf("%.9Lf\n", (long double)(0));
      continue;
    }
    long long ans = r == l ? 0 : (s[r] - s[l + 1]);
    ans *= (C.second - H);
    ans +=
        min(R * C.second,
            ((C.first - a[l].second) * H + a[l].second * C.second)) -
        max(L * C.second, ((C.first - a[l].first) * H + a[l].first * C.second));
    if (l != r)
      ans += min(R * C.second,
                 ((C.first - a[r].second) * H + a[r].second * C.second)) -
             max(L * C.second,
                 ((C.first - a[r].first) * H + a[r].first * C.second));
    printf("%.9Lf\n", ans / (long double)(C.second));
  }
}